STATISITICS AND PROBABILITY

HANDOUT
3rd QUARTER

RANDOM VARIABLE Example;

 Probability distribution of tossing a coin
 It is a way to assign a real number to each twice.
possible outcome in a random
event/experiment. X no. of heads P(X=x)
0 1/4 or 0.25
Example: 1 2/4 or 0.50
 Suppose our random variable X is the 2 1/4 or 0.25
number of heads in tossing a coin. Then, in
tossing a coin; Note: The sum of the probabilities in a discrete
probability distribution must be equal to 1.00
Possible Outcomes X no. of heads
Head 1
HOW TO IDENTIFY THE VALUES OF A
Tail 0
RANDOM VARIABLE?

To determine the values of any random

TYPES OF RANDOM VARIABLE variable, we need to identify first the list of all
possible outcomes (also known as sample space)
 Discrete random variable - its values are
of a random event and then define the random
obtained through counting.
variable in a way that we can assign a real number
 Continuous random variable - its values
to each possible outcome.
are obtained through measurement.
i) To generate the sample space in tossing a
Example: coin twice, we could use tree diagram. We
Type of use H to denote if it is a head and T if it is a
Random tail.
Variable TREE DIAGRAM
1. Number of honor
Discrete
students in a class
2. Growth of tomato
Continuous
seedlings per week
3. Time needed to finish an
Continuous
exam
4. Weight of pigs after 3
Continuous
months
5. Number of tiles needed
Discrete
to cover a floor
ii) To identify the values of the random
DISCRETE PROBABILITY DISTRIBUTION variable X, now that we have the sample
space in tossing a coin twice sample space
 A table which consists of the values a in tossing a coin twice, we define X as the
random variable can assume and the number of heads;
corresponding probabilities of the values. Possible Outcomes X no. of heads
 Also known as Probability Mass Function HH 2
HT 1
TH 1
TT 0
HOW TO ILLUSTRATE DISCRETE COMPUTING THE MEAN, VARIANCE,
PROBABILITY DISTRIBUTION? AND STANDARD DEVIATION OF A
DISCRETE PROBABILITY DISTRIBUTION
To illustrate the probability distribution of a
discrete random variable, we need to list down all Symbol
the determined unique values of the random Population Mean 𝛍
variable and compute the probabilities of each Population Variance 𝛔𝟐
random variable. Population Standard Deviation 𝛔

Example: Formulas
Suppose that a student will be playing 3 𝛍 = ∑ 𝐗 ∙ 𝐏(𝐗)
rounds of Rock-Paper-Scissors with another 𝛔𝟐 = ∑ 𝐗 𝟐 ∙ 𝐏(𝐗) − 𝛍𝟐
student. Illustrate the probability distribution of 𝛔 = √𝐯𝐚𝐫𝐢𝐚𝐧𝐜𝐞
the random variable X if X is the number wins.
Example:
Sample space X no. of wins Solve the mean, variance, and standard
WWW 3 deviation of the following probability
WWL 2 distribution.
WLW 2 1.
WLL 1 X P(X=x)
LWW 2 2 0.45
LWL 1 4 0.25
LLW 1 6 0.15
LLL 0 8 0.15
Probability Distribution of the Student Number of 𝛍 = ∑ 𝐗 ∙ 𝐏(𝐗)
Wins = 2(0.45) + 4(0.25) + 6(0.15) + 8(0.15)
= 4.00
X P(X=x)
0 1/8 or 0.125 𝛔𝟐 = ∑ 𝐗 𝟐 ∙ 𝐏(𝐗) − 𝛍𝟐
1 3/8 or 0.375 = (22(0.45)+42(0.25)+62(0.15)+82(0.15))
2 3/8 or 0.375 ─ 42
3 1/8 or 0.125 = 4.80
𝛔 = √𝐯𝐚𝐫𝐢𝐚𝐧𝐜𝐞
= √4.8
COMPUTING PROBABILITIES OF A ≈ 2.19
RANDOM VARIABLE WITH A DISCRETE
PROBABILITY DISTRIBUTION 2.
X P(X=x)
Example: 0 2/10
Using the probability distribution below, 1 3/10
X P(X=x) 2 3/10
4 0.15 3 2/10
5 0.45
6 0.25 𝛍 = ∑ 𝐗 ∙ 𝐏(𝐗)
7 0.15 = 0(2/10) + 1(3/10) + 2(3/10) + 3(2/10)
= 1.50
Compute the following probabilities;
1) P(X = 4) = 0.15 𝛔𝟐 = ∑ 𝐗 𝟐 ∙ 𝐏(𝐗) − 𝛍𝟐
2) P(X < 4) = 0 = (02(2/10) + 12(3/10) + 22(3/10)
3) P(X > 4) = 0.45 + 0.25 + 0.15 = 0.85 + 32(2/10)) ─ 1.52
4) P(X ≥ 4) = 0.15 + 0.45 + 0.25 + 0.15 = 1.05
= 1.00 𝛔 = √𝐯𝐚𝐫𝐢𝐚𝐧𝐜𝐞
5) P(X ≤ 5) = 0.15 + 0.45 = 0.60
= √1.05
≈ 1.02
BINOMIAL DISTRIBUTION REAL-LIFE PROBLEMS INVOLVING
RANDOM VARIABLES AND
 It is a probability experiment that satisfies PROBABILITY DISTRIBUTION
the following requirements:
1) Each trial have two possible outcomes Example:
or the number of outcomes can be 1) Suppose that the chances of the number
reduce to two outcomes only. of subjects that the students fail to pass
2) It has a fixed number of trials. in a year are;
3) The outcome of each trial must be x 0 1 2 3
independent with each other. P(X=x) 0.90 0.05 0.03 0.02
4) The probability of success is the same
for each trial. i) What are the chances that a student
will pass all of the subjects?
 It shows the probabilities of the possible P(X = 0) = 0.90
values of the random variable X where X is
Therefore, there is 90% that a
defined as the number of success achieved
student will pass all of the subjects for a
in a fixed number of trials in a binomial
whole school year.
experiment.
ii) What is the probability that a student
 Is defined as; P(X=x) = nCx px(1─p)n-x will at most fail 1 subject in a year?
Example: P(X ≤ 1) = 0.90 + 0.05 = 0.95
1) Suppose a coin will be toss 8 times. Therefore, the probability that a
What is the probability that the outcome student will at most fail 1 subject in a
will be exactly 5 heads? year is 0.95.
Given: n = 8
x=5 iii) What is the mean of the number of
p = ½ or 0.5 subjects that the students fail to pass in
a year?
Solution:
P(X=5) = 8C5 (0.5)5 (1 ─ 0.5)8-5 𝛍 = ∑ 𝐗 ∙ 𝐏(𝐗)
≈ 0.2188 = 0(0.90) + 1(0.05) + 2(0.03)
+ 3(0.02)
2) In a 10-item multiple choice exam, = 0.17
what is the probability of getting 8 Therefore, the average number of
correct items? subjects that the students fail to pass in
Given: n = 10 a year is 0.17.
x=8
p = ¼ or 0.25 2) In a basketball league, a particular team
only have 5 remaining games to be
Solution: played to end the season. Compute the
P(X=8) = 10C8 (0.25)8 (1 ─ 0.25)10-8 probabilities of the following;
≈ 0.0004
i) Win all the remaining games.
3) In a 5-item multiple choice exam, what P(X=5) = 5C5 (0.5)5 (1 ─ 0.5)5-5
is the probability of getting 1 incorrect ≈ 0.0313
item?
ii) Win only 3 games
Given: n = 5
x=1 P(X=5) = 5C3 (0.5)3 (1 ─ 0.5)5-3
p = ¾ or 0.75 = 0.3125
Solution: iii) Lose 4 games
P(X=1) = 5C1 (0.75)5 (1 ─ 0.75)5-1 P(X=4) = 5C4 (0.5)4 (1 ─ 0.5)5-4
≈ 0.0046 ≈ 0.1563
NORMAL DISTRIBUTION COMPUTING THE AREAS OR
PROBILITIES UNDER THE NORMAL
 It is a special type of probability density CURVE
curve that describes the tendency for data to
cluster at the center. To compute an area or probabilities under the
 Also known as Normal Curve or Bell normal curve, we should understand the different
Curve. probability notations, illustrate the areas to be
Illustration: computed, and look for the corresponding values
of the areas in Z-table.

PROBABILITY NOTATIONS

Probability Notation Area or Probability

P(Z < a) Table value
P(Z > a) 1 – Table value
Bigger value –
P(a < Z < b)
Smaller value
PROPERTIES OF NORMAL CURVE
Example:
 The distribution curve is bell-shaped. 1. P(Z < 1.75) = 0.9599
 The curve is symmetrical about its center.
 The mean, the median, and the mode
coincide at the center.
(Mean = Median = Mode)
 The parameters 𝛍 and 𝛔 completely
characterize the curve.
1.75
𝛍 : determines the position of the curve Using the z-table (preferably based
𝛔 : determines the spread or width of the from the left);
curve
 The total area under the curve is 1. z 0.00 ˑˑˑˑˑˑ 0.05
 The curve is asymptotic to the base line. 0.0
:
:
Z─TABLE 1.7 0.9599

 It is a table which displays all the areas of 2. P(Z < -2) = 0.0228
the regions under the standard normal curve
given a z-value (also known as Standard
score).
 Generally, there are two types of Z-table.
i.) Z-table based from the CENTER

-2
Using the z-table (preferably based
from the left);

z 0.00
ii) Z-table based from the LEFT -3.8
:
:
-2.0 0.0228
3. P(Z > 2.5) = 1 – 0.9938 = 0.0062 or CONVERTING RAW SCORE TO
= P(Z < -2.5) = 0.0062 STANDARD SCORE AND VICE VERSA

Important Symbols:
z : standard score
x : raw score
𝛍 : population mean
𝛔 : population standard deviation
2.5
Using the z-table (preferably based  To convert a raw score to standard score,
from the left); we use;

z 0.00 𝐱−𝛍
𝐳=
-3.8 𝛔
:
:  To convert a standard score to raw score,
-2.5 0.0062 we use;
x= 𝐳𝛔 + 𝛍
4. P(Z > -1.5) = 1 – 0.0668 = 0.9332 or
= P(Z < 1.5) = 0.9332
Example:

1. Convert the raw score x = 38 to standard

score if μ = 36 and σ = 4.

𝟑𝟖−𝟑𝟔
𝐳= = 𝟎. 𝟓
𝟒
-1.5
2. What is the corresponding standard
Using the z-table (preferably based
score of the raw score x = 31 if μ = 36
from the left);
and σ = 4?
z 0.00 𝟑𝟏−𝟑𝟔
0.0 𝐳= = −𝟏. 𝟐𝟓
𝟒
:
: 3. Convert the standard score z = 2.5 to
1.5 0.9332 raw score if μ = 36 and σ = 4.

5. P(-1.5 < Z < 1.5) = 0.9332 – 0.0668 𝐱 = 𝟐. 𝟓(𝟒) + 𝟑𝟔 = 𝟒𝟔

= 0.8664
4. Convert the standard score z = ─1.5 to
raw score if μ = 36 and σ = 4.

𝐱 = −𝟏. 𝟓(𝟒) + 𝟑𝟔 = 𝟑𝟎
5. What is the corresponding raw score of
-1.5 1.5 the standard score z = 1.75 if μ = 36
and σ = 4?
6. P(1.5 < Z < 1.75) = 0.9599 – 0.9332
= 0.0267
𝐱 = 𝟏. 𝟕𝟓(𝟒) + 𝟑𝟔 = 𝟒𝟑
6. What is the corresponding raw score of
the standard score z = ─15 if μ = 36
and σ = 4?

𝐱 = −𝟏(𝟒) + 𝟑𝟔 = 𝟑𝟐
1.5 1.75
REAL-LIFE PROBLEMS INVOLVING REFLECTION
NORMAL DISTRIBUTION
 What are your least mastered topics?
Example:

Suppose that the scores in a standardized

mathematics exam that is taken by 1,200
examinees were normally distributed. The
results revealed a mean of 34.00 with a
standard deviation of 5.00.  What are the topics you mastered?

1. What is the standard score of a student

with an actual score of 32 in the exam?

Solution:
𝟑𝟐−𝟑𝟒
𝐳= = −𝟎. 𝟒 Please rate the following statements based on the
𝟓
scale below.
2. What are the chances that a student
have a score of at least 40? 5 –Strongly Agree
4 –Agree
Solution: 3 –Undecided
2 –Disagree
P(X ≥ 40) = P(Z ≥ 1.2) 1 –Strongly Disagree
= 1 ─ 0.8849
= 0.1151
Score
Therefore, the chances that a student 1. The handout is easy to read.
have a score of at least 40 is 11.51% 2. The handout is comprehensive
3. The handout helped you to
3. What is the percentile rank of a student understand your least mastered
with a score 43? topics.
4. The handout engaged you to
Solution: study the subject more.
P(X < 43) = P(Z < 1.8) 5. The handout improved your
= 0.9641 confidence to take the quarter
exam.
Therefore, the percentile rank of a
student with a score 43 is 96.41%
Prepared by:
4. Approximately, how many students got
a score of 36 and above? CHRENIEL LOU G. ALECIDA, MS
SHS T-II
Solution:
P(X ≥ 36) = P(Z ≥ 0.4)
= 1 – 0.6554
= 0.3446
n = 0.3446 (1200)
≈ 414
Therefore, there are approximately 414
students with scores of 36 and above.